A factory produces weighted balls to use for exercise, by filling spherical rubber shells of different sizes with a sand-like material. The material's density is $1.5$ grams per cubic centimeter. Assuming the shell weighs $10$ grams, what should be the ball's radius so, when full, it weighs $1$ kilogram (or $1000$ grams)? Round your answer, if necessary, to the nearest tenth.
Explanation: This is a density word problem. To solve it, we can use the following equation, which is the volume definition of density: ${\text{Density}}=\dfrac{{\text{Total quantity}}}{{\text{Volume}}}$ What do we know? The ${\text{density}}$ of the filling material is ${1.5}$ grams per cubic centimeter. The ball's weight should be ${1000}$ grams. Once we subtract the shell's weight (which is $10$ grams), we get the ${\text{total quantity}}$. What do we need to find? The ball's radius, which gives us its ${\text{volume}}$. We want the ${\text{volume}}$, so let's rewrite the equation so ${\text{volume}}$ is isolated: $\begin{aligned} {\text{Density}}&=\dfrac{{\text{Total quantity}}}{{\text{Volume}}} \\\\ {\text{Density}}\cdot{\text{Volume}}&=\dfrac{{\text{Total quantity}}}{\cancel{{\text{Volume}}}}\cdot\cancel{{\text{Volume}}} \\\\ \dfrac{\cancel{{\text{Density}}}\cdot{\text{Volume}}}{\cancel{{\text{Density}}}}&=\dfrac{{\text{Total quantity}}}{{\text{Density}}} \\\\ {\text{Volume}}&=\dfrac{{\text{Total quantity}}}{{\text{Density}}} \end{aligned}$ As mentioned above, the actual ${\text{total quantity}}$ is $1000-10={990}$ grams, since we are trying to find the volume of the filling material within the shell. Let's denote the ball's radius as $r$. Then, the ${\text{volume}}$ is ${\dfrac43\pi r^3}$. Now we can plug ${\text{density}=1.5}$, ${\text{total quantity}=990}$, and ${\text{volume}=\dfrac43\pi r^3}$ in the equation. $\begin{aligned} {\text{Volume}}&=\dfrac{{\text{Total quantity}}}{{\text{Density}}} \\\\ {\dfrac43\pi r^3}&=\dfrac{{990}}{{1.5}} \\\\ r^3&=\dfrac{990}{1.5}\cdot\dfrac34\cdot\dfrac1\pi \\\\ r^3&=\dfrac{495}{\pi} \\\\ r&=\sqrt[3]{\dfrac{495}{\pi}} \\\\ &\approx 5.4 \end{aligned}$ The radius of the ball should be approximately $5.4$ centimeters.